Problem 72 - College Algebra with Modeling and Visualization

Question

Solve each equation. Use the change of base formula to approximate exact answers to the nearest hundredth when appropriate. $$3 \cdot 5^{x}=125$$

Step-by-Step Solution

Answer

The solution is \(x \approx 2.43\).

Step 1: Isolate the Exponential Expression

To solve the equation \(3 \cdot 5^x = 125\), we first need to isolate the term involving the exponent. This is done by dividing both sides of the equation by 3:\[5^x = \frac{125}{3}\]

Step 2: Apply the Change of Base Formula

To solve for \(x\), we will use the logarithm. The change of base formula is \(\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\). In this case, we choose base 10 for \(k\) because it's commonly used in calculators:\[x = \frac{\log(\frac{125}{3})}{\log(5)}\]

Step 3: Calculate Using a Calculator

Calculate the values of the logarithms using a calculator:1. Calculate \(\log(\frac{125}{3})\).2. Calculate \(\log(5)\).3. Divide the result from step 1 by the result from step 2 to find \(x\).Make sure your calculator is in the correct mode for calculating logarithms. This gives:\[x \approx \frac{1.69897}{0.69897} \approx 2.43\]

Step 4: Round to the Nearest Hundredth

The final step is to round your answer from Step 3 to the nearest hundredth. Therefore:\[x \approx 2.43\]

Key Concepts

Exponential Equations

An exponential equation is a type of equation where a variable appears in an exponent. In math, these equations occur in various scenarios, such as calculating compound interest or population growth. Exponential equations have a specific form, generally written as \(a \cdot b^x = c\), where \(b\) is the base and \(x\) is the exponent in the equation. These equations often require solving through specialized methods, as seen in the example using the equation \(3 \cdot 5^x = 125\). To approach these problems, the goal is to first isolate the exponential term for easier manipulation. Neutralizing complicating terms, such as multiplying constants or additional addends, often helps to focus solely on the exponential component. Once isolated, understanding logarithmic operations becomes essential in determining the value of the variable \(x\).

Change of Base Formula

The change of base formula is a valuable tool for solving exponential equations. It is particularly useful when the base of the exponent doesn’t match the base used in calculations. The formula is written as:

• \(\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\)

This allows us to switch the base \(b\) to another base \(k\) that is more convenient for computation, typically base 10 or base \(e\) (natural logarithm). For \((3 \cdot 5^x = 125)\), after isolating \((5^x)\), we apply the formula by choosing base 10. This choice is practical for most calculators, simplifying calculation by transforming the problem into solvable algebraic steps. The change of base formula effectively converts difficult exponential terms into expressions manageable through logarithms, making it indispensable in solving complex equations.

Logarithms

Logarithms are the inverse operations of exponentiation. When you take the logarithm of a number, you are essentially asking, "To what power must a specific base be raised to give this number?" The formula \(b^x = a\) translates to \(x = \log_b(a)\). For solving exponential equations, particularly when isolating \(x\), logarithms simplify the process. Let's consider solving \(5^x = \frac{125}{3}\). We can rewrite this using logarithms:

• \(x = \log_5(\frac{125}{3})\)

Here, utilizing the change of base formula bridges the transition from exponentials to logarithmic expressions. With the base 10 conversion, calculators effortlessly evaluate these expressions, making solving seemingly complex exponential equations straightforward. This transformation is crucial for isolating variables in the exponent and providing exact or approximate numerical solutions.

Calculator Usage

Using a calculator is essential for calculating logarithms, especially when dealing with non-standard bases. Modern calculators allow users to input both the base and the argument of a logarithm, but most computations rely on base 10 or natural logs (base \(e\)). For example, to solve \(\log_5(\frac{125}{3})\), employ a calculator by first calculating \(\log_{10}(\frac{125}{3})\) and \(\log_{10}(5)\).

• Enter \(\frac{125}{3} = 41.67\) to calculate \(\log_{10}(41.67)\)
• Next, calculate \(\log_{10}(5)\)
• Divide these values to find \(x\)

Ensure the calculator is in base 10 mode, as accurate computation hinges on the correct settings. Calculators simplify these processes, ensuring fast and precise results, ideal for solving advanced algebraic equations like exponential functions. Their use underscores the practicality of mathematical computation in education, reinforcing theoretical concepts through practical application.